By the end you can
Learn $\sin 2\theta$, the three forms of $\cos 2\theta$, and $\tan 2\theta$.
Rearrange them to write $\cos^2\theta$ and $\sin^2\theta$ without squares.
Find exact values such as $\cos 22\tfrac12^\circ$, and prove identities.
Put $\alpha = \beta = \theta$ in the compound-angle formulae.
Why it works
The second and third forms of $\cos 2\theta$ come from replacing $\sin^2\theta$ by $1 - \cos^2\theta$, or $\cos^2\theta$ by $1 - \sin^2\theta$. Choose the form that leaves only one trig function in your problem.
Squares in terms of the double angle
Rearranging the last two forms of $\cos 2\theta$:
$$\cos^2\theta = \tfrac12(1 + \cos 2\theta), \qquad \sin^2\theta = \tfrac12(1 - \cos 2\theta).$$
cos²x as a wave Turn on JavaScript to use this interactive graph.
Orange: y = cos²x. Blue dashed: y = ½(1 + cos 2x), exactly the same curve. Faint pink: y = cos 2x. So cos²x is a cosine wave with half the period, squashed into the band from 0 to 1.
Exact values
$\sin\theta = \dfrac35$ and $\theta$ is obtuse. Find $\sin 2\theta$, $\cos 2\theta$ and $\tan 2\theta$.
$\cos\theta = -\dfrac45$ and $\tan\theta = -\dfrac34$.
Obtuse, so cosine is negative.
$\sin 2\theta = 2 \times \dfrac35 \times \left(-\dfrac45\right) = -\dfrac{24}{25}$
$\cos 2\theta = 1 - 2 \times \dfrac{9}{25} = \dfrac{7}{25}$, so $\tan 2\theta = -\dfrac{24}{7}$.
Example 2: a half angle ●●● Find the exact value of $\cos 22\tfrac12^\circ$.
$\cos^2 22\tfrac12^\circ = \tfrac12(1 + \cos 45^\circ) = \tfrac12\left(1 + \dfrac{\sqrt2}{2}\right) = \dfrac{2 + \sqrt2}{4}$
Use cos²θ = ½(1 + cos 2θ) with θ = 22½°.
$\cos 22\tfrac12^\circ = \dfrac{\sqrt{2 + \sqrt2}}{2}$
22½° is acute, so take the positive root.
Proving identities
Prove that $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$.
LHS $= \dfrac{1 - (1 - 2\sin^2 x)}{2\sin x\cos x} = \dfrac{2\sin^2 x}{2\sin x\cos x}$
Use the 1 − 2sin² form so the 1s cancel.
$= \dfrac{\sin x}{\cos x} = \tan x = $ RHS
Example 4: a triple angle ●●● Show that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$.
$\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta\cos\theta + \cos 2\theta\sin\theta$
$= 2\sin\theta\cos^2\theta + (1 - 2\sin^2\theta)\sin\theta$
$= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta$
Replace cos²θ by 1 − sin²θ.
Common mistake
$\sin 2\theta$ is not $2\sin\theta$. At $\theta = 90^\circ$: $\sin 180^\circ = 0$, but $2\sin 90^\circ = 2$.
$sin 2 heta$
$2\sin\theta\cos\theta$
$cos 2 heta$ (three forms)
$\cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$
$ an 2 heta$
$\dfrac{2\tan\theta}{1 - \tan^2\theta}$
$sin^2 heta$
$\tfrac12(1 - \cos 2\theta)$
$\sin 2\theta = $
$2\sin\theta$ $2\sin\theta\cos\theta$ $\sin^2\theta - \cos^2\theta$ $\sin\theta\cos\theta$ From $\sin(\theta + \theta)$.
Which is not equal to $\cos 2\theta$?
$\cos^2\theta - \sin^2\theta$ $2\cos^2\theta - 1$ $1 - 2\sin^2\theta$ $2\sin^2\theta - 1$ That one is $-\cos 2\theta$.
$\sin\theta = \dfrac35$, $\theta$ obtuse. Find $\sin 2\theta$.
$2 \times \dfrac35 \times \left(-\dfrac45\right)$.
$\sin\theta = \dfrac35$. Find $\cos 2\theta$.
$1 - 2 \times \dfrac{9}{25}$. It does not depend on the sign of $\cos\theta$.
$\tan\theta = -\dfrac34$. Find $\tan 2\theta$.
$\dfrac{-\frac32}{1 - \frac{9}{16}} = -\dfrac{24}{7}$.
$\cos^2\theta = $
$\tfrac12(1 + \cos 2\theta)$ $\tfrac12(1 - \cos 2\theta)$ $2\cos 2\theta - 1$ $1 + \cos 2\theta$ From $\cos 2\theta = 2\cos^2\theta - 1$.
Evaluate $2\sin 15^\circ\cos 15^\circ$.
Evaluate $1 - 2\sin^2 30^\circ$.
$\dfrac{1 - \cos 2x}{\sin 2x}$ simplifies to:
$\sin x$ $\cot x$ $\tan x$ $1$ $\sin 3\theta = $
$3\sin\theta$ $3\sin\theta - 4\sin^3\theta$ $4\sin^3\theta - 3\sin\theta$ $3\sin\theta\cos\theta$ Try again
New numbers every time. Keep going until you get five in a row.
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